(x-4)(3x+2)=x^2-20x+4

Solution for (x-4)(3x+2)=x^2-20x+4 equation:

(x-4)(3x+2)=x^2-20x+4

We move all terms to the left:

(x-4)(3x+2)-(x^2-20x+4)=0

We get rid of parentheses

-x^2+(x-4)(3x+2)+20x-4=0

We multiply parentheses ..

-x^2+(+3x^2+2x-12x-8)+20x-4=0

We add all the numbers together, and all the variables

-1x^2+(+3x^2+2x-12x-8)+20x-4=0

We get rid of parentheses

-1x^2+3x^2+2x-12x+20x-8-4=0

We add all the numbers together, and all the variables

2x^2+10x-12=0

a = 2; b = 10; c = -12;
Δ = b2-4ac
Δ = 102-4·2·(-12)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-14}{2*2}=\frac{-24}{4}
=-6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+14}{2*2}=\frac{4}{4}
=1
$